Question

A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating ∫ i2R dt and also by finding the decrease in the energy stored in the capacitor.

Answer 1

Answer:

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Answer 2

Explanation:

Let the capacitor’s initial charge be Q0. Then,

Q0 = CV

The capacitor’s charge at time t after making the connections,

Q = Q0e - tRC

i = dQdt = -Q0RCe-tRC

During time t1 to t2, the dissipated heat is

U = Q022Ce - 2t1RC -e -2t2RC

Time constant = RC

Putting t2 = RC and t1 = 0, we obtain:

U = Q022Ce-0-e-2

Therefore, Q0 = CV, U = CV21 - 1e2

Alternative method:

Heat discharged at any time = − Energy deposited at time t + Energy deposited at time 0

⇒U = 12CV2 - 12CV2e - 2tRC

Therefore, t = RC, ⇒U = 12CV21 – 1e2